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I'm stuck on the following probability problem and would welcome any help:

Consider three random uniform variables on [0,1]. Let X be the minimum of these three variables and Y be the maximum. What is the expected value of X*Y?

Am I right in assuming I need to find the joint distribution function of X and Y? I have found the separate distribution functions but I'm having trouble with the joint one since X and Y are not independent.

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    $\begingroup$ And... what did you try to find the joint distribution of $(X,Y)$? $\endgroup$ – Did Jan 22 '19 at 17:15
  • $\begingroup$ The distribution function for X is: $P(X<x) = 1-(1-x)^3$. Likewise, $P(Y<y) = y^3$. So for the joint function, $P(X<x, Y<y)$ I tried separating the cases $x>y$ and $x<y$. Indeed, if $x>y$ then, $P(X<x,Y<y) = P(Y<y) = y^3$ since in this case $X<Y<y<x$. For the case $x<y$ I'm kind of lost, but it is perhaps possible that here X and Y are independent, in which case we have $P(X<x,Y<y) = y^3*(1-(1-x)^3)$. If you can help find the joint function I'd be very grateful, although I now realise that passing through joint functions is not the easiest solution. $\endgroup$ – nicdel Jan 23 '19 at 0:11
  • $\begingroup$ $X$ and $Y$ are certainly not independent, because both are concentrated on $[0,1]$ but $X \le Y$ always. $\endgroup$ – Robert Israel Jan 23 '19 at 1:21
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It's simpler to deal with the three uniform random variables $U_1, U_2, U_3$. Given $U_1 < U_2 < U_3$, $XY = U_1 U_3$. So $$\eqalign{ \mathbb E [XY | U_1 < U_2 < U_3] &= \mathbb E[U_1 U_3 | U_1 < U_2 < U_3]\cr &= 6 \int_0^1 du_3 \int_0^{u_3} du_2 \int_0^{u_2} du_1 \; u_1 u_3}$$ By symmetry, the result is the same for all the other orderings of $U_1, U_2, U_3$, so this is also $\mathbb E[XY]$.

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